AP EAPCET 2024 - 22th May Evening Shift | Motion in a Straight Line Question 3 | Physics

A particle starts from rest and moves in a straight line. It travels a distance $2 L$ with uniform acceleration and then moves with a constant velocity a further distance of $L$. Finally, it comes to rest after moving a distance of $3 L$ under uniform retardation. Then, the ratio of average speed to the maximum speed $\left(\frac{v}{v_m}\right)$ of the particle is

$\frac{6}{11}$

$\frac{7}{11}$

$\frac{5}{11}$

$\frac{2}{11}$

AP EAPCET 2024 - 21th May Evening Shift

MCQ (Single Correct Answer)

-0

The velocity of a particle is given by the equation $v(x)=3 x^2-4 x$, where $x$ is the distance covered by the particle. The expression for its acceleration is

$(6 x-4)$

$6\left(3 x^2-4 x\right)$

$\left(3 x^2-4 x\right)(6 x-4)$

$(6 x-4)^2$

AP EAPCET 2024 - 21th May Evening Shift

MCQ (Single Correct Answer)

-0

The acceleration of a particle which moves along the positive $X$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \mathrm{~ms}^{-1}$ at $x=0$ , then its velocity at $x=1.4 \mathrm{~m}$ is $\left(\right.$ in $\left.\mathrm{ms}^{-1}\right)$

AP EAPCET 2024 - 21th May Evening Shift Physics - Motion in a Straight Line Question 4 English

1.6

1.2

1.4

0.8

AP EAPCET 2024 - 21th May Morning Shift

MCQ (Single Correct Answer)

-0

An object projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of 8 s and the object reaches foot after 16 s . The height of the tower is $\left(g=10 \mathrm{~ms}^{-2}\right)$

220 m

240 m

640 m

80 m

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